Given: #n^2-3n-40=0#
#color(blue)("General observation (reading the equation)")#
As the #n^2# term is positive the graph is of generic shape #uu#
The #n# term is negative so the the vertex is to the right of the y-axis thus #n_("vertex")>0#
As the vertex is to the right of the y-intercept then the #y_("vertex") < y_("intercept")# and the graph 'crosses the n-axis.
Thus there are two solutions to #n^2-3n-40=0#
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#color(blue)("Deriving the determinant and thus the count of solutions")#
Consider the standard form #0=y=ax^2+bx+c#
where #x=(-b+-sqrt(b^2-4ac))/(2a)#
The determinant part is #b^2-4ac#
In this case: #a=+1; b=-3 and c=-40# giving:
#b^2-4ac ->(-3)^2-4(1)(-40) = 169#
As the determinant is greater than 0 it also tells us that
#color(brown)(ul("there are 2 solutions."))#
